Mixed Numbers: Free Response

The two readings are two different calculations, so carry both out in full before comparing them.

Under the mixed-number convention the space means add, so the whole number joins the fraction. Write the $3$ in sevenths, since one whole is $\tfrac{7}{7}$ and three wholes are $3 \times 7 = 21$ sevenths:

$3\tfrac{2}{7} = \frac{21}{7} + \frac{2}{7} = \frac{23}{7}.$

Read the same space as a multiplication and the whole number scales the fraction instead, which is the multiply-across rule with $3$ written as $\tfrac{3}{1}$:

$3 \times \frac{2}{7} = \frac{3}{1} \times \frac{2}{7} = \frac{6}{7}.$

Both calculations are correct arithmetic. What separates them is not the arithmetic but the instruction the notation is understood to carry, and the two results are nothing like each other.

Place each value by asking how many times the denominator fits inside the numerator.

For the mixed-number reading, seven fits into twenty-three three times with two sevenths left over, so the value has passed $3$ but has not reached $4$:

$3 < \frac{23}{7} < 4.$

For the multiplication reading, seven does not fit into six at all, so the value has not even reached one whole:

$0 < \frac{6}{7} < 1.$

Now read the notation the way a reader does. The $3$ at the front announces three complete wholes and a little more. Only one of the two values delivers that. The addition reading sits in the interval the leading $3$ points at, while the multiplication reading sits below $1$ and leaves the $3$ describing nothing at all.

That is what makes the convention worth its oddity. A notation whose leading number told you nothing about the size of the value would be harder to read than the improper fraction it replaced, and the whole point of a mixed number is that its size can be seen at a glance.

The safeguard is the condition a mixed number always satisfies: its fraction part is proper, so it is less than $1$, and its whole part is at least $1$.

Take the multiplication reading first, because its inequality needs nothing checked. Multiplying by a number below $1$ takes only part of a quantity, and the whole part is at least $1$, so the product lands strictly below the whole number:

$3 \times \frac{2}{7} < 3 \times 1 = 3.$

The addition reading can never go below the whole number, since a fraction part is never negative, and it goes strictly above as soon as that fraction part is above zero, which it is in every mixed number anyone writes:

$3 \le 3 + \frac{2}{7}.$

So the two readings land on opposite sides of the number written at the front: the multiplication reading strictly below it, the addition reading never below it. A reader who glances at the leading number knows at once which side the value is meant to be on, and the convention costs nothing in practice because the rival reading is never plausible. (The single case where the addition reading lands exactly on the whole number is a fraction part of zero, such as $3\tfrac{0}{9}$, and nobody writes that, because it is just $3$.)

Now drop the condition and watch the safeguard fail. Suppose someone writes $2\tfrac{5}{3}$, with a fraction part larger than $1$. The addition reading gives $2 + \tfrac{5}{3} = \tfrac{11}{3}$ and the multiplication reading gives $2 \times \tfrac{5}{3} = \tfrac{10}{3}$:

$\frac{11}{3} > 2 \quad \text{and} \quad \frac{10}{3} > 2.$

Both land above the $2$, and they sit only $\tfrac{1}{3}$ apart besides, so nothing about the size of the result tells a reader which was meant. That is why a mixed number is required to carry a proper fraction. The requirement is not tidiness; it is what keeps the notation readable.

The fraction bar was a division before it was anything else, so ask the plain division question: how many complete groups of $8$ fit inside $53$, and what is left standing outside them?

$53 \div 8 = 6 \text{ remainder } 5.$

Now read those two numbers back through the fraction. Eight eighths make one whole, so each complete group of $8$ eighths is one whole unit: the quotient $6$ counts wholes. The remainder $5$ is the eighths that could not be gathered into another whole, so it counts leftover eighths, and it keeps the denominator it was already measured in:

$\frac{53}{8} = 6\tfrac{5}{8}.$

The leftover is guaranteed to be a proper fraction, and that is not luck. A remainder is always smaller than the number you divided by, so the remainder is always smaller than the denominator, which is exactly the condition a mixed number's fraction part has to meet.

Take them one at a time, and simplify only once the division is done.

Six fits into $34$ five times, using $30$, and leaves $4$:

$34 \div 6 = 5 \text{ remainder } 4, \qquad \frac{34}{6} = 5\tfrac{4}{6}.$

The fraction part is not yet in lowest terms, since $4$ and $6$ share the factor $2$, so divide top and bottom by it:

$5\tfrac{4}{6} = 5\tfrac{2}{3}.$

The second conversion behaves differently. Nine fits into $72$ exactly eight times and leaves nothing over:

$72 \div 9 = 8 \text{ remainder } 0, \qquad \frac{72}{9} = 8.$

There is no fraction part to write. The conversion does not record a zero remainder at all, so what it hands you is the whole number $8$. Writing $8\tfrac{0}{9}$ instead names the same value and breaks no rule, but it is not the output of the conversion: the fraction part exists to record what was left over, and here nothing was. Either way the standard result is the whole number $8$, which is a reminder that an improper fraction does not always convert into a whole number with a fraction beside it.

A claim of the form "exactly when" is two claims facing opposite ways, and each needs its own test.

Start with the direction that works. Suppose a fraction converts to a mixed number: a whole number $w$ of at least $1$ with a proper fraction $\tfrac{a}{b}$ beside it. A fraction part is never negative, so the value is at least the whole part:

$w + \frac{a}{b} \ge w \ge 1.$

A fraction of whole numbers whose value is at least $1$ has a numerator at least as large as its denominator, and that is exactly what improper means, so this direction is sound. It is $\ge 1$ that does the work here rather than $> 1$: a value of exactly $1$ is still improper, as $\tfrac{9}{9}$ shows below.

Now the other direction, and part B has already supplied the case that settles it. The fraction $\tfrac{72}{9}$ is improper, since its numerator is far larger than its denominator, and yet the conversion leaves no remainder at all:

$\frac{72}{9} = 8.$

The result is a whole number with nothing standing beside it, so the classmate's test would refuse to call $\tfrac{72}{9}$ improper. It is improper. One case is enough to sink a direction.

The boundary is worth naming too, because it fails the same way and is easy to miss. When the numerator equals the denominator,

$\frac{9}{9} = 1,$

the fraction is improper under the usual definition, and again it converts to a whole number with no fraction part.

The repair is to stop testing by the shape of the converted form and go back to the definition: for a fraction of whole numbers, it is improper exactly when the numerator is greater than or equal to the denominator. That version holds in both directions, and it covers the whole-number cases the classmate's version leaves out.

"One and a half times as much" is a multiplication, so the length wanted is $1\tfrac{1}{2}$ of the standard bed's $3\tfrac{2}{3}$ feet. Neither number can be multiplied in mixed form, so convert both first: multiply the whole part by the denominator, add the numerator, and keep the denominator.

$1\tfrac{1}{2} = \frac{1 \times 2 + 1}{2} = \frac{3}{2}, \qquad 3\tfrac{2}{3} = \frac{3 \times 3 + 2}{3} = \frac{11}{3}.$

Now multiply across, cancelling the $3$ that stands on the top of one fraction and the bottom of the other:

$\frac{3}{2} \times \frac{11}{3} = \frac{3 \times 11}{2 \times 3} = \frac{11}{2}.$

The answer is wanted as a mixed number, so divide: $11 \div 2 = 5$ remainder $1$.

$\frac{11}{2} = 5\tfrac{1}{2}.$

A large bed takes $5\tfrac{1}{2}$ feet of strip. The size is believable, and a second route confirms it: half as much again as $3\tfrac{2}{3}$ feet means adding $1\tfrac{5}{6}$ feet to it, and $3\tfrac{4}{6} + 1\tfrac{5}{6} = 4\tfrac{9}{6} = 5\tfrac{1}{2}$.

Dividing asks how many copies of one length fit inside another, which is exactly what the question wants. Convert both mixed numbers first:

$16\tfrac{1}{2} = \frac{33}{2}, \qquad 3\tfrac{2}{3} = \frac{11}{3}.$

Dividing by $\tfrac{11}{3}$ means multiplying by its reciprocal $\tfrac{3}{11}$:

$\frac{33}{2} \div \frac{11}{3} = \frac{33}{2} \times \frac{3}{11}.$

Cancel before multiplying, since $33$ and $11$ share the factor $11$ and leave $3$ on top:

$\frac{33}{2} \times \frac{3}{11} = \frac{3 \times 3}{2} = \frac{9}{2} = 4\tfrac{1}{2}.$

So one roll goes $4\tfrac{1}{2}$ of the way through the standard beds. Reporting $\tfrac{9}{2}$ would be the same value, but the mixed form is the one that shows at a glance how many beds are covered.

A quotient is counted in whatever unit the division was asking about, and the division asked how many standard beds fit. So the quotient is measured in beds. Its whole part, $4$, says four complete beds can be edged, and its fraction part, $\tfrac{1}{2}$, says that after those four there is enough strip for half of another bed. It is not half of a foot.

To find the strip actually left, work in feet. Four beds use

$4 \times 3\tfrac{2}{3} = \frac{4}{1} \times \frac{11}{3} = \frac{44}{3} = 14\tfrac{2}{3} \text{ feet}.$

Take that away from the roll, over the common denominator $6$:

$\frac{33}{2} - \frac{44}{3} = \frac{99}{6} - \frac{88}{6} = \frac{11}{6} = 1\tfrac{5}{6} \text{ feet}.$

So $1\tfrac{5}{6}$ feet of strip remain on the roll.

The two leftovers differ because they count different things: one counts beds and the other counts feet. They agree perfectly once the unit is converted, and the check is a single multiplication, half of one bed's length:

$\frac{1}{2} \times \frac{11}{3} = \frac{11}{6} = 1\tfrac{5}{6}.$

Reading the fraction part of a quotient in the wrong unit is the standard trap in a division like this one. The cure is to ask what the division was counting before reading its answer back into the situation.

Each mixed number is a sum, so the whole calculation is a sum of four terms:

$1 + \frac{7}{10} + 2 + \frac{4}{5}.$

Addition is commutative and associative, so those four terms may be reordered and bracketed however you like without changing the total. Gather the wholes together and the fractions together:

$(1 + 2) + \left(\frac{7}{10} + \frac{4}{5}\right).$

The wholes give $3$. For the fractions, the least common denominator of $10$ and $5$ is $10$, so rewrite the second fraction and add:

$\frac{7}{10} + \frac{8}{10} = \frac{15}{10} = \frac{3}{2} = 1\tfrac{1}{2}.$

The fraction column has passed one whole, and that is the wrinkle. The extra whole cannot stay in the fraction column, because a mixed number's fraction part must be proper, so carry it across:

$3 + 1\tfrac{1}{2} = 4\tfrac{1}{2}.$

Stopping at $3\tfrac{15}{10}$ would name the same value, but it is not in mixed-number form and it hides the size of the number, which is the one job the form exists to do.

Put the fraction parts over a common denominator first. The least common denominator of $9$ and $3$ is $9$, and $\tfrac{2}{3} = \tfrac{6}{9}$, so the subtraction reads

$7\tfrac{2}{9} - 3\tfrac{6}{9}.$

Now the trouble is visible: $\tfrac{6}{9}$ cannot be taken from $\tfrac{2}{9}$ and leave a fraction part behind. Exchange one of the seven wholes for its worth in ninths. One whole is $\tfrac{9}{9}$, so the $7$ becomes $6$ and the fraction column gains nine ninths:

$7\tfrac{2}{9} = 6 + \frac{9}{9} + \frac{2}{9} = 6\tfrac{11}{9}.$

Nothing has changed in value, and the reason is visible on that line: the whole given up and the ninths gained are the same quantity, because one whole is $\tfrac{9}{9}$. The same number is simply written with a different split between the two columns, which is all an exchange ever does. Now subtract column by column:

$6\tfrac{11}{9} - 3\tfrac{6}{9} = (6 - 3) + \left(\frac{11}{9} - \frac{6}{9}\right) = 3 + \frac{5}{9} = 3\tfrac{5}{9}.$

The fraction part $\tfrac{5}{9}$ is proper, and $5$ and $9$ share no factor above $1$, so $3\tfrac{5}{9}$ is finished.

Start with the size check, because it costs nothing and it settles the matter on its own. Over a common denominator the subtraction is $7\tfrac{2}{9} - 3\tfrac{6}{9}$. The fraction being taken away is the larger of the two, so the four wholes left by $7 - 3$ cannot survive intact: part of one of them has to go as well. The result is therefore below $4$:

$7\tfrac{2}{9} - 3\tfrac{6}{9} < 7 - 3 = 4.$

The student reports $4\tfrac{4}{9}$, which is above $4$, so the report is wrong before a single careful step is taken.

Now the diagnosis. The step that fails is the fraction column, where the two fractions were swapped to avoid a subtraction that looked impossible. Order matters in a subtraction, so the swap does not compute the same thing:

$\frac{6}{9} - \frac{2}{9} = \frac{4}{9}, \qquad \text{while the line calls for} \qquad \frac{2}{9} - \frac{6}{9}.$

That column was supposed to take $\tfrac{4}{9}$ away from the four wholes; instead it added $\tfrac{4}{9}$ to them. Going the wrong way by $\tfrac{4}{9}$ costs that amount twice, so the miss is

$\frac{4}{9} + \frac{4}{9} = \frac{8}{9},$

and the two answers confirm it, since $4\tfrac{4}{9} - 3\tfrac{5}{9} = \tfrac{40}{9} - \tfrac{32}{9} = \tfrac{8}{9}$.

The repair is the exchange from part B. Move one whole into the fraction column and the subtraction there becomes genuinely possible, so nothing has to be turned round.

Convert each factor by multiplying the whole part by the denominator, adding the numerator, and keeping the denominator:

$2\tfrac{1}{5} = \frac{11}{5}, \qquad 1\tfrac{1}{2} = \frac{3}{2}.$

Multiply across, then convert back with a division, $33 \div 10 = 3$ remainder $3$:

$\frac{11}{5} \times \frac{3}{2} = \frac{33}{10} = 3\tfrac{3}{10}.$

The shortcut does something else entirely. It multiplies the whole parts, $2 \times 1 = 2$, and separately the fraction parts,

$\frac{1}{5} \times \frac{1}{2} = \frac{1}{10},$

and sets the two results side by side as $2\tfrac{1}{10}$. The two answers are more than a whole apart, so this is not a small slip in the arithmetic. The shortcut is computing a different quantity, and the next part shows which one.

Each mixed number is a sum, so this product is a product of two sums:

$2\tfrac{1}{5} \times 1\tfrac{1}{2} = \left(2 + \frac{1}{5}\right)\left(1 + \frac{1}{2}\right).$

The distributive property takes a single outside factor across a sum, $a(b + c) = ab + ac$, so put the second bracket in that outside position. Multiplication may be taken in either order, so swap the two factors first:

$\left(2 + \frac{1}{5}\right)\left(1 + \frac{1}{2}\right) = \left(1 + \frac{1}{2}\right)\left(2 + \frac{1}{5}\right).$

Now the rule applies in the form it was taught, with $a = 1 + \tfrac{1}{2}$ outside the sum $2 + \tfrac{1}{5}$:

$\left(1 + \frac{1}{2}\right)\left(2 + \frac{1}{5}\right) = 2\left(1 + \frac{1}{2}\right) + \frac{1}{5}\left(1 + \frac{1}{2}\right),$

where each product has again been written with its single factor first. Apply the rule a second time inside each of those two pieces:

$2\left(1 + \frac{1}{2}\right) = 2 \times 1 + 2 \times \frac{1}{2}.$

$\frac{1}{5}\left(1 + \frac{1}{2}\right) = \frac{1}{5} \times 1 + \frac{1}{5} \times \frac{1}{2}.$

Those are the four products. Two of them are the ones the shortcut formed: $2 \times 1$, the whole parts, and $\tfrac{1}{5} \times \tfrac{1}{2}$, the fraction parts. The other two each pair one number's whole part with the other number's fraction part, and the shortcut never formed them at all.

Add all four and the correct product appears, with the fractions over the common denominator $10$:

$2 + 1 + \frac{1}{5} + \frac{1}{10} = 3 + \frac{2}{10} + \frac{1}{10} = 3\tfrac{3}{10}.$

The two missing cross products come to $1 + \tfrac{1}{5} = 1\tfrac{1}{5}$, which is exactly the gap between the shortcut's report and the true product.

Write both mixed numbers out and gather the like pieces:

$2\tfrac{1}{5} + 1\tfrac{1}{2} = 2 + \frac{1}{5} + 1 + \frac{1}{2} = (2 + 1) + \left(\frac{1}{5} + \frac{1}{2}\right) = 3 + \frac{7}{10} = 3\tfrac{7}{10}.$

Nothing was thrown away there. All four terms are still present; they were only reordered and rebracketed, which is exactly what the commutative and associative properties of addition permit. That is the whole reason the column method is exact for a sum: separating whole parts from fraction parts is a regrouping, and addition does not care how a sum is grouped.

Multiplication does care. Part B showed what a product of two sums actually contains: four products, not two. The part by part shortcut keeps the two that pair like with like and silently drops the two cross products, and a term that has been dropped does not come back. Multiplication distributes across a sum rather than ignoring it, so the two operations behave quite differently on the thing a mixed number is, namely a sum.

One condition remains on the addition side, and it is a condition on the form of the result rather than on the method. The regrouping is always valid, but what the columns hand back is not always in mixed-number form yet: when the fraction parts reach one whole or more, the fraction column is improper, and that whole has to be carried into the whole-number column before the answer counts as a mixed number. Here $\tfrac{1}{5} + \tfrac{1}{2} = \tfrac{7}{10}$ is proper, so nothing carries. Change the numbers so the fraction parts pass one whole and the carry becomes necessary, but even then no term is ever lost, which is the difference that matters.